# Vector valued function derivative example | Multivariable Calculus | Khan Academy

What I want to do in this videos is to make to
parametrizations of essentially the same curve, but we're going
to go along the curve a different rates.
And hopefully we'll be able to use that to understand, or get
a better intuition, behind what exactly it means to take a
derivative of a position vector valued function.
So let's say my first parametrization, I have
x of t is equal to t.
And let's say that y of t is equal to t squared.
And this is true for t is greater than or equal to 0,
and less than or equal to 2.
And if I want to write this as a position vector valued
function, let me write this.
x1, call that y1, and let me write my position vector valued
function; I could say r1-- I'm numbering them because I'm
going to do a different version of this exact same curve with a
slightly different parametrization --so r1 one of
t, we could say is x1 of t times i-- the unit vector i
--so we'll just say t times i plus-- this is just x of t
right here, or x1 of t; I'm numbering them because I'll
later have an x2 t --plus t squared times j.
And if I wanted to graph this, I'm going to be very careful
graphing it because I really want to understand what the
derivative means here.
Try to draw it roughly to scale.
So let's say that this is one, two, three, four.
Then let me draw my x-axis.
That's good enough.
And my x-axis, I want it to be roughly to scale, one and two.
And so at t equals 0, both my x and y coordinates are at 0-- or
this is just going to be the 0 vector, so this is where we are
a t equals 0 --at t equals 1 this is going to be one times
i-- we're going to be just like that --plus 1 times j.
1 squared is j, so we're going to be right there.
And then at t is equal to 2, we're going to be at 2i.
So 2i-- you could imagine 2 times i would be this vector
right there --2 times i plus 4-- 2 squared is 4 --4 times
j, so plus 4 times j.
to get a vector that's end point is right there.
The vector is going to look something like this.
So this is what, just to make it clear what we're
doing, that's r1 of 2.
This is r1 of 0.
This is r1 of 1.
But the bottom line is the path looks like this:
it's a parabola.
So the path will look like that.
Now that's in my first parametrization of it.
Actually, let me draw a little bit more carefully.
I want to get rid of this arrows, just because I want it
to be a nice clean drawing.
So it's going to be a parabola.
Let me get rid of that other point, too, just because I
didn't draw it exactly where it needs to be; it needs
to be right there.
And my parabola, or part of my parabola is going to
look something like that.
All right.
Good enough.
So this is the first parametrization.
Now I'm going to do this exact same curve, but I'm going to
do it slightly differently.
So let's say I'll do it in different colors.
So x2 of t, let's it equals 2t.
And y2 of t, let's say it's equal to 2t squared.
Or we could alternatively write that, that's the same thing as
4t squared, just phrasing both of these guys
to the second power.
And then let's say instead of going from t equals 0 to 2,
we're going to go from t goes from 0 to 1.
But we're going to see, we're going to cover
the exact same path.
And our second position vector valued function, r2 of t, is
going to be equal to 2t times i plus-- I could say 2t squared
--4t squared times j.
And if I were to graph this guy right here, it would look
like-- let me draw my axes again; it's going to look the
same, but it's I think useful to draw it because I'm going to
draw the derivatives and all that on it later.
One, two, three, four.
One, two.
And then let's see what happens when t is equal to 0-- or r of
0; all these are going to be 0, we're just going to have the
zero vector; x and y are both equal to 0 --when t is equal
to 1/2 what are we going to get here?
1/2 times 2 is 1.
And then we're going to get the point 1/2 squared
is 1/4 times 4 is 1.
So when t is equal 1/2 we're going to be at the point 1, 1.
And when the t is equal to 1 we're going to
be at the point 2, 4.
So notice the curve is exactly, the path we go
is exactly the same.
But before we even do the derivatives, these two
paths are identical.
I want to think about something.
Let's pretend that our parameter, t, really is time.
And that tends to be the most common, that's
why they call it t.
It doesn't have to be time, but let's say it is time.
So what's happening here?
In the first parametrization when we go from 0 to 2
seconds we cover this path.
You can imagine after 1 second the dot moves here,
then it moves there.
You can imagine a dot moving along this curve, and it
takes two seconds to do so.
In this situation we have a dot moving along the same curve,
but it's able to cover the same curve in only one second; and
half a second it gets here.
It took this guy one second to get here.
In a one second, this guy's all the way over here; this guy
takes two seconds to go over here.
So in this second parametrization even though the
path is the same, the curves are the same, the
dot is faster.
I want you to keep that in mind when we think about
the derivatives of both of these position vector
valued functions.
So just remember the dot is moving faster for every second
it's getting further along the curve than here; that's why it
only took them one second.
Now let's look at the derivatives of both
of these guys.
So the derivative here, so if I were to write r prime, r1 prime
of t-- let me do that in a different color, actually,
already used the orange; so let me do it in the blue
--r1 prime us t.
So the is the derivative now.
It's going to be, remember, it's just the derivative of
each of these times the unit vectors.
So the derivative of t with respect to t, that's just 1.
So it's 1 times i.
I'll just write 1i plus-- I didn't have to write the one
there --plus the derivative of t squared with respect
to t is 2t plus 2t j.
And let me take the derivative over here.
r2 prime of t.
The derivative of 2t with respect to t is 2, so
2i, plus the derivative of 4t squared is 8t.
2 times 4, it is rt.
Just like that.
Now the question is, what do their respective derivative
vectors look like at different points?
So let's look at, I don't know, let's see how fast they're
moving when time is equal to 1.
So let's take it at a specific point.
This is just the general formula, but let's figure
out what the derivative is at a specific point.
So let's take r1 when time is equal to 1.
And I want to take this specific point on the curve,
not the specific point in time.
So this point on the curve here is when time is equal
to 1, you could say second.
This point over here, which is the exact corresponding
point, is when time is equal 1/2 second.
So r1 of 1 is equal to-- we're taking the derivative
there --is equal to 1i.
It's not dependent on t at all.
So it's 1i plus 2 times 1j, so plus 2j.
So at this point the derivative of our position vector function
is going to be 1i plus 2j.
So we can draw it like this. so if we do 1i is like this: 1i.
And then 2j.
Just 2j is like that.
So our derivative right there, I'll do it in the same
color that I wrote it in.
It's in this green color; it's going to look like this.
And notice it looks like, at least its direction is-- let me
do it a little bit straighter --its direction looks tangent
to the curve; it's going in the direction that my
particle is moving.
Remember my particle is moving from here to there, so it's
going in the direction.
And I'm going to think about, in a second, what this length
of this to derivative vector is.
This right here, just to be clear is, r1 prime.
It's a vector, so it's telling us the instantaneous change in
our position vector with respect to t, or time, when
time is equal to 1 second.
That's this thing right here.
Now let's take the exact same position here on our curve.
But that's going to occur at a different time for this guy.
We already said it only takes him, he's here at time
is equal to 1/2 second.
So let's take-- --I'll do it in the same color
--so here we have r2.
We're going to evaluate it at 1/2 half because this is at
time is equal 1/2 second.
And this is going to be equal to 2i-- this isn't dependent
at all on time --so 2i plus 8 times the time.
So time right here is 1/2.
So 8 times 1/2 is 4.
So plus 4j.
So what does this look like?
The instantaneous derivative here.
Oh, and this is the derivative; have to be very clear.
So 2i-- let me draw some more --so 2i maybe gets
Plus 4j will get us up to right around there.
Plus 4j is that factor.
So when you add those two heads to tails, you get this thing:
you get something that-- let me like --you get something
that looks like that.
I didn't draw it as neatly as I would like to.
But let's notice something: both of these vectors are going
in the exact same direction.
They're both tangential to the path, to our curve.
But this vector is going, its length, its magnitude, is
much larger than this vector's magnitude.
And that makes sense because I hinted at it when we first
talked about these vector valued position functions and
their derivatives; is that the length, you can kind of
view it as the speed.
The length is equal to the speed if you imagine t being
time and these parametrizations are representing a dot
moving along these curves.
So in this case, the particle only takes a second to go
there, so at this point in its path, it's moving much faster
than this particle is.
So if you think about it, this vector right here, if you
imagine this is a position factor, this is velocity.
Velocity is speed plus the direction.
Speed is just you know, how fast are you going?
Velocity is how fast you're going in what direction?
I'm going this fast-- and you could calculate it using the
Pythagorean Theorem, but I just want to give you the intuition
right here --I'm going that fast in this direction.
Here I'm going this fast; I'm going even faster.
That's my magnitude, but I'm still going in
the same direction.
So hopefully you have a gut feeling now of what the
derivative of these position vectors really are.